Question

A study indicates that teenagers spend an average of 112 minutes watching videos on their smartphones per week. Assume the distribution is normal, with a standard deviation of 12 minutes. What is the probability that a teenager spends less than 90 minutes watching videos on their phone per week

Answer 1

Answer:

0.0336 = 3.36% probability that a teenager spends less than 90 minutes watching videos on their phone per week.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

A study indicates that teenagers spend an average of 112 minutes watching videos on their smartphones per week. Assume the distribution is normal, with a standard deviation of 12 minutes.

This means that $\mu = 112, \sigma = 12$

What is the probability that a teenager spends less than 90 minutes watching videos on their phone per week?

This is the p-value of Z when X = 90. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{90 - 112}{12}$

$Z = -1.83$

$Z = -1.83$ has a p-value of 0.0336

0.0336 = 3.36% probability that a teenager spends less than 90 minutes watching videos on their phone per week.